Methyl tert-butyl ether (MTBE), C5H12O, a gasoline additive used ...

Question

Methyl tert-butyl ether (MTBE), C5H12O, a gasoline additive used to boost octane ratings, has ΔH°f = -313.6 kJ/mol. Write a balanced equaiton for its combustion reaciton, and calcualte its standard heat of combustion in kilojoules

Accepted Answer

Hi everyone for this problem, it reads what is the balanced equation for the combustion reaction of isopropyl ether? A gasoline additive that has a standard entropy a formation equal to negative 318 kill the jewels per mole. And what is its standard heat of combustion and kill a jewels? So for this problem, we have two things that we want to answer here. The first is the balanced equation for the combustion reaction, and the second is its standard heat of combustion. So let's go ahead and start with part one. Okay, so for combustion reaction, it's a reaction with oxygen producing hydrogen and carbon dioxide. So we have our di isopropyl ether and this is a liquid and it's going to react with oxygen gas. Two produce hydrogen and carbon dioxide. Okay, so we have H two a liquid plus carbon dioxide gas. So we're going to do what we're going to do here is we're going to count our reactant and products. Okay, so we have carbon, we have hydrogen and we have oxygen. So we're going to go ahead and write out what we have for react ints and product. Okay, so for our reactant, we have six carbons, We have 14 hydrogen and three oxygen's for our products. There is one carbon, two hydrogen and three oxygen's. Okay, so what we're going to want to do here is we're going to want to multiply our carbons by six and our hydrogen is by seven. Okay, so we're going to take this and we're going to multiply this by six and multiply this by seven so that they match. So when we rewrite this now, we're going to get the following. Okay, so what we're going to do is we're going to go ahead and count our reactant and products again to make sure that they match. Okay, so we have carbon, hydrogen and oxygen react tints and products. So now we see that we have in our reactant since six carbons 14 hydrogen and three oxygen's. And for our products, there are six carbons hydrogen and oxygen's. And we want this number. We want these numbers to match. Okay, so what we're going to do here is we're going to multiply our oxygen's by nine. Okay, we're gonna put a nine in front of our oxygen's so that the coefficient match. Okay, so once we do that, we're going to get a balanced reaction is going to be the following. Okay, so this is our balanced reaction. And the second part of the problem asks us to calculate the standard heat of combustion in order for us to calculate the standard heat of combustion. We're going to need to write down the standard heats of formation for all that's involved in our reaction. So these are values that we're going to need to look up. So let's go ahead and write those values down the standard heat of formation for our di isopropyl ether, This is equal to negative 318 killer jewels per mole. Our standard heat a formation for carbon dioxide gas is equal to negative five kila jewels per mole. Our standard heat of formation for H 2 liquid is equal to negative 285.8 kilo joules per mole. And our standard heat of formation for oxygen gas is equal to zero kg joules per mole because anything in its elemental state is equal to zero. So the equation that we're going to use to solve for standard heat of formation or standard entropy change. Excuse me. Which is what we're Yes, our standard heat of combustion, this is going to equal the some of the products minus the sum of the reactant. So we're going to take the number of moles we have from our balanced equation. And we're going to multiply it by its standard heat of formation. Okay, and so what that is going to look like is we're gonna plug in those values. So it's going to equal the sum of our products. So for our products we have H 20 liquid and carbon dioxide gas. So we're going to take the standard heat of formation for or let's write it this way. So what we're going to do is we're going to take the number of moles. So we have we have for our products and this balanced equation should have an arrow instead of a plus. So this is our correct balanced equation here. Okay, so it's going to equal the sum of our products, minus the sum of our reactant. So we have seven moles of H 20 liquid. So we're going to take seven moles and multiply it by its standard heat of formation. So for H 20 liquid, it is negative 285.8 kg joules per mole plus we have six moles of carbon dioxide, so six most times the standard heat of formation for carbon dioxide is negative 393.5 kg joules per mole. So that's the sum of the products. And we're going to subtract it by the sum of the reactant. So for our reactant we have one mole of Diana's, Oh, propose ether. So we're going to write one here and it's standard PT formation is negative 318 killer jewels Permal. And this is gonna be plus, we have one mole of or we have nine moles of oxygen gas, but it's standard heat of formation is zero killer jewels per mole. So essentially we're going to just ignore this. Alright, so that is the equation that we're going to use to solve for our standard entropy of combustion or standard heat of combustion. And once we do this calculation, we're going to get a final answer of negative 4043. kg joules per mole. And this is going to be our final answer. So we have our balanced equation for the combustion reaction as well as the standard heat of combustion and kill a jules. That's it for this problem, I hope this was helpful.

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